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# Inequality 25(Generalization)

The following problem is a generalization of a well known inequality:

Problem:

Let $\displaystyle a,b,c$ be positive real numbers such that $\displaystyle ab+bc+ca=3$. Prove that

$\displaystyle \frac{1}{(a+kb)^3}+\frac{1}{(b+kc)^3}+\frac{1}{(c+ka)^3}\geq \frac{3}{(k+1)^3}$, where $\displaystyle k$ is a non-negative real number.

Solution:

From Holder’s inequality we get that

$\displaystyle \left[\sum_{cyc}\frac{1}{(a+kb)^3}\right]\left[\sum_{cyc}a(b+kc)\right]^{3}\geq \left(\sum_{cyc}a^{\frac{3}{4}}\right)^{4}$, or $\displaystyle \sum_{cyc}\frac{1}{(a+kb)^3}\geq \frac{\left(a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}\right)^{4}}{(k+1)^3(ab+bc+ca)^3}$.

So, it suffices to prove that

$\displaystyle \frac{\left(a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}\right)^{4}}{(k+1)^3(ab+bc+ca)^3}\geq \frac{3}{(k+1)^3}$.

But the last relation is equivalent to $\displaystyle \left(\sum_{cyc}a^{\frac{3}{4}}\right)^{4}\geq 81=9\sqrt{3}(ab+bc+ca)^{3/2}$.

Let us denote by $\displaystyle x,y,z$ the $\displaystyle a^{3/4},b^{3/4},c^{3/4}$ respectively. Then $\displaystyle ab=(xy)^{4/3},bc=(yz)^{4/3},ca=(zx)^{4/3}$.

So, our inequality takes the form $\displaystyle (x+y+z)^4\geq 9\sqrt{3}\sum_{cyc}(xy)^{4/3}$. This inequality is homogeneous so, we consider the sum $\displaystyle x+y+z$ equal to $\displaystyle 3$.

Doing some manipulations in left and right hand side we only need to prove that

$\displaystyle \sum_{cyc}(xy)^{4/3}\leq 3$.

Now from the AM-GM inequality we have that

$\displaystyle \sum_{cyc}xy\cdot (xy)^{1/3}\leq \sum_{cyc}xy\frac{x+y+1}{3}=\sum_{cyc}xy\frac{4-z}{3}$.

The last one is equal to $\displaystyle \sum_{cyc}(xy)^{4/3}\leq \frac{4}{3}\sum_{cyc}xy-xyz$.

After that, we only need to prove

$\displaystyle \frac{4}{3}\sum_{cyc}xy-xyz\leq 3\Longrightarrow 4\sum_{cyc}x\sum_{cyc}xy\leq 27+9xyz=\left(\sum_{cyc}x\right)^{3}+9xyz$,

which is Schur’s 3rd degree inequality, Q.E.D.